Let \( \displaystyle R \) be a finite boolean ring. Show that there exists a set \( \displaystyle X \) and a ring isomorphism \( \displaystyle R \cong \mathcal{P}(X) \).
Finite Boolean Ring
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Finite Boolean Ring
Here is a really difficult exercise!
Let \( \displaystyle R \) be a finite boolean ring. Show that there exists a set \( \displaystyle X \) and a ring isomorphism \( \displaystyle R \cong \mathcal{P}(X) \).
Let \( \displaystyle R \) be a finite boolean ring. Show that there exists a set \( \displaystyle X \) and a ring isomorphism \( \displaystyle R \cong \mathcal{P}(X) \).
HINT
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Re: Finite Boolean Ring
Hi Nickos.
Firstly, if \(\displaystyle{X}\) is a non-empty set, then, the power set \(\displaystyle{\mathbb{P}(X)}\)
is a \(\displaystyle{\rm{Boolean}}\) ring with addition
\(\displaystyle{A+B:=(A-B)\cup(B-A)\,,\forall\,A\,,B\in\mathbb{P}(X)}\)
and multiplication
\(\displaystyle{A\cdot B:=A\cap B\,,\forall\,A\,,B\in\mathbb{P}(X)}\)
and then \(\displaystyle{0_{\mathbb{P}(X)}=\varnothing\,\,\,,1_{\mathbb{P}(X)}=X}\).
Nickos, my idea is to define the map
\(\displaystyle{\Phi:R\to \mathbb{P}(\rm{Hom}(R,\mathbb{Z}_{2}))\,,r\mapsto \Phi(r)=\left\{f\in\rm{Hom}(R,\mathbb{Z}_{2}), f(r)=1\right\}}\)
which is a ring homomorphsim.
I am not sure if \(\displaystyle{\Phi}\) is one to one and onto.
I like very much this question and i gave an answer in order to make some progress.
Firstly, if \(\displaystyle{X}\) is a non-empty set, then, the power set \(\displaystyle{\mathbb{P}(X)}\)
is a \(\displaystyle{\rm{Boolean}}\) ring with addition
\(\displaystyle{A+B:=(A-B)\cup(B-A)\,,\forall\,A\,,B\in\mathbb{P}(X)}\)
and multiplication
\(\displaystyle{A\cdot B:=A\cap B\,,\forall\,A\,,B\in\mathbb{P}(X)}\)
and then \(\displaystyle{0_{\mathbb{P}(X)}=\varnothing\,\,\,,1_{\mathbb{P}(X)}=X}\).
Nickos, my idea is to define the map
\(\displaystyle{\Phi:R\to \mathbb{P}(\rm{Hom}(R,\mathbb{Z}_{2}))\,,r\mapsto \Phi(r)=\left\{f\in\rm{Hom}(R,\mathbb{Z}_{2}), f(r)=1\right\}}\)
which is a ring homomorphsim.
I am not sure if \(\displaystyle{\Phi}\) is one to one and onto.
I like very much this question and i gave an answer in order to make some progress.
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